GATE 2025 IN Q.33: solution and answer

Official answer: 2Source: Official answer key, GATE 2025 IN (IIT Roorkee)

1. Factor the denominator: $z^2-z=z(z-1)$, so $f(z)=\frac{2z+1}{z(z-1)}$. The only singular points are $z=0$ and $z=1$. The numerator $2z+1$ is not zero at either point (it equals $1$ and $3$), so both are simple poles.

2. At a simple pole $z=a$ the residue is ${\mathrm{Res}}_{z=a}f={\lim }_{z\to a}(z-a)f(z)$.

3. At $z=0$: ${\mathrm{Res}}_{z=0}f={\frac{2z+1}{z-1}|}_{z=0}=\frac{1}{-1}=-1$.

4. At $z=1$: ${\mathrm{Res}}_{z=1}f={\frac{2z+1}{z}|}_{z=1}=\frac{3}{1}=3$.

5. Sum of the residues: $-1+3=2$.

6. Check with partial fractions: $\frac{2z+1}{z(z-1)}=\frac{-1}{z}+\frac{3}{z-1}$, because $\frac{-(z-1)+3z}{z(z-1)}=\frac{2z+1}{z(z-1)}$. The residues are the coefficients, $-1$ and $3$.

7. Second check: for large $|z|$, $f(z)=\frac{2z+1}{z^2-z}\approx \frac{2}{z}$, so the residue at infinity is $-2$. All residues, including the one at infinity, add to zero, so the finite residues must add to $+2$.